Problem: Is ${134971}$ divisible by $3$ ?
A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {134971}= &&{1}\cdot100000+ \\&&{3}\cdot10000+ \\&&{4}\cdot1000+ \\&&{9}\cdot100+ \\&&{7}\cdot10+ \\&&{1}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {134971}= &&{1}(99999+1)+ \\&&{3}(9999+1)+ \\&&{4}(999+1)+ \\&&{9}(99+1)+ \\&&{7}(9+1)+ \\&&{1} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {134971}= &&\gray{1\cdot99999}+ \\&&\gray{3\cdot9999}+ \\&&\gray{4\cdot999}+ \\&&\gray{9\cdot99}+ \\&&\gray{7\cdot9}+ \\&& {1}+{3}+{4}+{9}+{7}+{1} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${134971}$ is divisible by $3$ if ${ 1}+{3}+{4}+{9}+{7}+{1}$ is divisible by $3$ Add the digits of ${134971}$ $ {1}+{3}+{4}+{9}+{7}+{1} = {25} $ If ${25}$ is divisible by $3$ , then ${134971}$ must also be divisible by $3$ ${25}$ is not divisible by $3$, therefore ${134971}$ must not be divisible by $3$.